The number of atoms in 4.5 g of a face-centred cubic crystal with edge length 300 pm is (Given : Density =10 ~g ~cm ⁻³ and .N_A=6.022 10²³ )

Step-by-step solution

Given, Face centred crystal, so Z=4 edge length (a)=300 pm =300 10⁻¹⁰ ~cm density (d)=10 ~g / cm ^3N_A=6.022 10²³ Using the formula d= Z M N_A a^3 M= d N_A a^3 Z = 10 6.022 10²³ (300 10⁻¹⁰ )^3 4 M=40.65 ~g So, 40.65 g contains 6.022 10²³ atoms Let 4.5 g contains x atom x= 6.022 10²³ 40.65 4.5=6.6 10²² atoms